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Find the equation of the line tangent to the curve $y=x^2$ parallel to the line $y=x$.

Just started A level maths, any help is appreciated.

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  • $\begingroup$ The slope of the line $y=x$ is $1$. That should be obvious. We must find where the slope of the curve $y=x^2$ is equal to $1$. Do you know how to find the slope of a curve? $\endgroup$ – Emily Oct 16 '14 at 19:13
  • $\begingroup$ Yes, I know how to find the slope of a curve. Thanks $\endgroup$ – Joe Oct 16 '14 at 19:15
  • $\begingroup$ Ok, so what is the equation that describes the slope of the curve $y=x^2$? $\endgroup$ – Emily Oct 16 '14 at 19:22
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The slope of $y=x$ is $1$.

We ask ourselves the following question: in what point is the tangent at $y=x^2$ $1$?

$y'=2x=1$ gives $x=\frac12$.

$y(\frac12)=\frac14$, so the line passing through $(x,y)=(\frac12,\frac14)$ with slope $1$ is $y=x-\frac14$.

enter image description here

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  • $\begingroup$ Thanks very much, helped a lot! Does this method apply to all questions of this type? $\endgroup$ – Joe Oct 16 '14 at 19:16
  • $\begingroup$ Yes, it applies to all questions of this type. $\endgroup$ – rae306 Oct 16 '14 at 19:17
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    $\begingroup$ Thank you very much for your help. $\endgroup$ – Joe Oct 16 '14 at 19:19
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Find $y'(x)$. When does $y'(x) = 1\;?$ ($m = 1$ is the slope of the line $y = x$, and hence the slope of any line parallel to $y = x$).

There will be one solution: name it $x_0$.

Find $y$ at $x_0$, and call it $y_0$.

Then you have the slope of the desired line, and the point $(x_0, y_0)$ and can use the point slope form of the equation of a line: $$y - y_0 = m(x-x_0)$$

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